Electronics E-Note for secondary schools

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Edwin Ogie Library — Electronics E-Note

Comprehensive lesson: Conductors • Semiconductors • Diodes • Transistors • Basic Circuits

Introduction — Why Electronics?

Electronics is the study and use of electrical devices that control the flow of electrons to process information, amplify signals, and convert energy. This note covers the basic solid-state building blocks used in modern circuits and practical examples you can try conceptually.

We'll follow a path from material classification, to p-n junctions and diodes, then to transistors as amplifiers and switches, finishing with basic passive circuits and worked problems.

Conductors, Insulators & Semiconductors (Band Theory)

In solids, electrons occupy energy bands. The two most important bands are:

• Valence band: occupied by electrons bound to atoms.
• Conduction band: where electrons are free to move and conduct current.

The band gap (Eg) is the energy separating these bands. Classification:

• Conductor: no band gap, bands overlap (metals).
• Insulator: large Eg (> ~3–4 eV) so electrons cannot jump easily.
• Semiconductor: moderate Eg (≈0.1–2 eV). Conductivity is controllable.

Temperature and conductivity

In semiconductors, raising temperature increases carrier concentration (more electrons jump the gap) — conductivity increases with temperature (opposite of metals).

Example — Comparing materials

Solution: Copper: conduction band and valence band overlap → conductor. Silicon: Eg ≈1.1 eV → semiconductor. Glass Eg > 4 eV → insulator.

Intrinsic & Extrinsic Semiconductors (Doping)

Intrinsic: pure semiconductor; carriers arise from thermal excitation (electrons and holes in equal numbers).

Extrinsic: deliberately doped to add carriers:

• n-type: dopant has extra valence electron (e.g., P in Si) → free electrons (majority).
• p-type: dopant has fewer valence electrons (e.g., B in Si) → holes (majority).

Why doping works

Adding a donor (pentavalent) introduces energy levels slightly below the conduction band; only small energy is needed to free the electron — dramatically increasing conductivity.

Example — Doping effect

Solution: Phosphorus (5 valence electrons) in Si bonds with 4 Si atoms; one electron remains free → n-type.

P-N Junctions & Diodes

A p-n junction forms where p-type and n-type semiconductors meet. At the junction:

• Electrons diffuse from n to p and recombine with holes — leaving behind fixed ions and a depletion region.
• A built-in potential (barrier) prevents further net diffusion at equilibrium.

Diode I–V characteristic (idealised)

When forward biased (positive on p side), the barrier lowers and current flows. When reverse biased, current is very small until breakdown.

I ≈ I₀ (e^{qV/(kT)} − 1) (Shockley diode equation — idealised)

Diode types & uses

• Signal diodes: rectification and small-signal switching.
• Zener diodes: voltage regulation (operate in breakdown safely).
• Schottky diodes: low forward voltage for fast switching.
• LEDs: emit light when forward biased.

Example — Diode rectifier (half wave)

Solution: A diode in series with AC will conduct during positive halves and block negative halves — output is pulsating DC with average ≈ Vp/π (for ideal diode and large load resistance approximation).

Example — Zener regulator

Setup: Input Vs > Vz, series resistor R, Zener diode to ground. The diode clamps voltage to Vz across load.

Solution: Choose R so that Iz (Zener current) stays between Iz(min) and Iz(max) across expected load variation: R = (Vs − Vz)/(Iz + IL) where IL is load current. Ensure power rating of Zener not exceeded.

Transistors — BJT & FET Basics

Transistors are three-terminal devices that amplify or switch. Two common families:

• BJT (Bipolar Junction Transistor): current-controlled device (base current controls collector current). Types: NPN, PNP.
• FET (Field Effect Transistor): voltage-controlled device (gate voltage controls channel current). Types: MOSFET, JFET.

BJT operating regions

• Cutoff: transistor off (like open switch).
• Active: used for linear amplification (collector current ≈ β × base current).
• Saturation: transistor fully on (like closed switch).

Small-signal (common-emitter) amplifier idea

A small change in base current produces a larger change in collector current; with a collector resistor this produces a larger voltage swing at the collector — amplification.

Example — BJT as switch

Scenario: Drive a 200 mA LED from 5 V using an NPN transistor. The transistor's β ≈ 100, VCE(sat)≈0.2 V.

Solution: To force saturation choose base current Ib ≈ Ic/10 = 200mA/10 = 20mA. If driven from a microcontroller pin (max 20 mA), instead use a base resistor from 5V via driver or use a MOSFET. Calculate resistor: Rb=(Vdrive−Vbe)/Ib ≈ (5−0.7)/0.02 ≈ 215 Ω. Also ensure collector resistor or LED series resistor sets Ic to 200 mA.

Example — Small-signal amplifier gain

Given: Rc=4.7kΩ, collector quiescent current Ic≈1mA, β≈100. Small-signal transconductance gm ≈ Ic/VT where VT≈25mV → gm≈1mA/25mV=0.04 A/V. Voltage gain Av≈−gm×Rc ≈ −0.04×4700 ≈ −188. (Large; real circuits set bias so gain and headroom are safe.)

Passive Components: Resistors, Capacitors & RC Circuits

Resistors

Resistors limit current and form voltage dividers: Vout = Vin×R2/(R1+R2).

Capacitors

Capacitors store charge Q=CV; reactance Xc=1/(2πfC). They block DC and pass AC (depending on frequency).

RC time constant

τ = R × C

Charging: Vc(t) = V(1−e^{−t/τ}); discharging: Vc(t) = V e^{−t/τ}.

Example — RC time constant

Given: R = 10 kΩ, C = 10 µF. τ = 10,000 × 10×10⁻⁶ = 0.1 s. At t = τ, Vc ≈ 0.632 × Vin.

Practical Design Tips & Rules of Thumb

• Always verify power ratings (P = I²R) for resistors and dissipation for diodes/transistors.
• Place decoupling capacitors (0.1 µF) close to IC power pins to bypass switching noise.
• For stability, bias transistors with proper resistor networks (avoid relying on β alone).
• Use proper pull-up/pull-down resistors for logic inputs to avoid floating states.

Example — Choosing base resistor

Scenario: Microcontroller pin (3.3 V) drives BJT to saturate at Ic=100mA. Assume Vbe≈0.7V, choose Ib=Ic/10 =10mA. Rb=(3.3−0.7)/10mA≈260Ω. Use next standard value 220Ω or 270Ω and check pin drive capability.

Worked Problems — Click to Reveal Solutions

Problem 1: A diode conducts when forward biased by 0.7 V (silicon). If a diode is in series with a 1kΩ resistor and a 9 V supply, calculate the current.

V_R = 9 − 0.7 = 8.3 V → I = V_R / 1kΩ = 8.3 mA.

Problem 2: A 10 µF capacitor charges through a 100 kΩ resistor from 12 V. Find τ and voltage after 3τ.

τ = R×C = 100000×10×10⁻⁶ = 1 s. After 3τ (~3 s) Vc ≈ 0.95×Vin = 11.4 V.

Problem 3: A load draws 200 mA at 12 V. Using an NPN transistor as a low-side switch with β ~ 100, what minimum base current ensures saturation? If driven from a 5 V logic pin (Vbe≈0.7), find base resistor value.

Ib ≈ Ic/10 = 20 mA (forced β ≈ 10 for saturation). Rb=(5−0.7)/0.02 ≈ 215 Ω. Check microcontroller pin current limits — this is near max; better use driver transistor or MOSFET.

Problem 4: For a voltage divider R1=10k, R2=5k from 12 V, what is Vout? If a load of 10k is connected across R2, find new Vout.

Open-circuit Vout = 12×5k/(10k+5k) = 4 V. With load RL=10k in parallel with R2: R2||RL = (5k×10k)/(5k+10k)=3.333k. Vout = 12×3.333/(10+3.333)=12×3.333/13.333≈3.0 V.

Problem 5: A BJT amplifier has Rc=4.7k and collector current Ic≈2 mA. Estimate small-signal voltage gain (approx −β×Rc/re'). Assume re'≈25mV/Ie and β≈100.

Ie≈Ic=2mA → re'≈25mV/2mA≈12.5 Ω. Av ≈ −β×Rc/(β×re' + re' approx) ~ −Rc/(re') ≈ −4700/12.5 ≈ −376. Real circuits use emitter resistors to stabilize and reduce gain to practical values.

Problem 6: Calculate the reactance of a 10 µF capacitor at 50 Hz.

Xc = 1/(2πfC) = 1/(2π×50×10×10⁻6) ≈ 318 Ω.

Problem 7: A Schottky diode has forward voltage ≈0.3 V. With 5 V supply and 330 Ω resistor, what is forward current?

I = (5 − 0.3)/330 ≈ 14 mA.

Problem 8: An LED requires 20 mA and forward drop 2.0 V. If driven from 12 V through resistor R, find R.

R = (12 − 2.0)/0.02 = 10/0.02 = 500 Ω. Use 470 Ω or 511 Ω standard value; check power rating.

Problem 9: Using an op-amp in voltage follower mode, what is the closed-loop gain?

Voltage follower: Vout = Vin, so gain = 1. Used for buffering high-impedance sources.

Problem 10: Determine the DC current through a 1k resistor connected to 9 V when a 0.7 V diode in series is present and the diode is forward biased.

V_R = 9 − 0.7 = 8.3 V → I = 8.3 mA.

Further Reading & Practical Tasks

• Try building simple diode and RC circuits on a breadboard; measure with a multimeter.
• Practice biasing transistors using a voltage divider and emitter resistor to stabilize operating point.
• Use SPICE (e.g., LTspice) to simulate transistor amplifiers and observe waveforms.

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